Total Accepted: 87074
Total Submissions: 348588
Difficulty: Hard
Contributors: Admin
Given an unsorted integer array, find the first missing positive integer.For example,
Given [1,2,0] return 3,
and [3,4,-1,1] return 2.Your algorithm should run in O(n) time and uses constant space.
方法:
遍历一遍,把正确合法的正整数方在正确的位置上即可。比如我们找到元素5,就把5换到A[4]上,因为数组从0下标开始。最终结果数组A[i] != i+1 的就是该位置上的正整数没有出现过,返回i+1即可。
代码如下:
class Solution(object):
def firstMissingPositive(self, nums):
n = len(nums)
for i in range(0, n):
while nums[i] > 0 and nums[i] <= n and nums[nums[i]-1] != nums[i]:
nums[nums[i]-1], nums[i] = nums[i], nums[nums[i]-1]
for i in range(0, n):
if nums[i] != i+1:
return i+1
return n+1
呵呵,人生苦短,我用Python。